Vague convergence of measures Define a subprobability measure to be a measure on Borel sigma algebra of the real line $\mathbb{R}$  with the measure for the whole real line less or equal than 1.
I was wondering about the definition of vague convergence of a sequence of subprobability measures $\{ \mu_n, n\geq 1 \}$ to another subprobability measure $\mu$. The convergence can be defined in slightly two different ways as in Chung's probability theory book p85 and p90:
(1) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b]) \rightarrow \mu((a,b])$.
(2) if there exists a dense subset D of the real line $\mathbb{R}$  so that $ \forall a \text{ and } b \in D \text{ with } a <b, \mu_n((a,b)) \rightarrow \mu((a,b))$ (the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?).
How to show these two definitions are equivalent?
A side question: 
is there a definition of vague convergence for general measures on more general sigma algebra with more general underlying space?
Thank you so much!
 A: Here is the general notion of vague convergence, taken from Olav Kallenberg's Foundations of Modern Probability (2nd edition).
Let $S$ be a locally compact, second countable Hausdorff space equipped with its Borel $\sigma$-field ${\cal S}$. Let $\hat{\cal S}$ be the ring of relatively compact Borel sets, and ${\cal M}(S)$ the space of locally finite non-negative measures.
Locally finite means that $\mu(B)<\infty$ when $B\in \hat{\cal S}$. 
The vague topology on ${\cal M}(S)$ is the topology generated by the mappings 
$\mu\mapsto \int f\ d\mu$ for every $f$ a non-negative continuous function with compact support.  
A: In my previous answer, I have shown that definition (1) implies definition (2). Now I prove the converse implication.
Suppose that the condition in definition (2) holds. That is, $\mu_n ((a,b))$ converges for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $\mathbb{R}$. We are now going to specify the limit. For any $a,b \in D$, $a < b$, define
$$
\mu ((a,b)) = \mathop {\lim }\limits_{n \to \infty } \mu _n ((a,b)).
$$
Since $\mu_n$ is a sequence of subprobability measures (s.p.m.), $\mu ((a,b)) \in [0,1]$.
Next, fix any $a,b \in {\mathbb R}$, $a < b$. If $a_m$ is a decreasing sequence converging to $a$, and $b_m$ is an increasing sequence converging to $b$, $a_m,b_m \in D$, then, by the last definition, $\mu ((a_m,b_m))$ is a monotone increasing sequence, bounded from above by $1$. Hence, this sequence converges, and we can define
$$
\mu ((a,b)) = \mathop {\lim }\limits_{m \to \infty } \mu ((a_m ,b_m )) \in [0,1].
$$
This actually determines a s.p.m. $\mu$ on $\mathbb{R}$. Now, define $\tilde D = \lbrace x \in {\mathbb R}:\mu (\lbrace x \rbrace ) = 0 \rbrace$ and proceed similarly to my previous answer, based on 
$$
\mu _n ((a',b')) \le \mu _n ((a,b]) \le \mu _n ((a'',b'')),
$$
to conclude that for any $a < b$ in (the dense set) $\tilde D$, $\mu _n ((a,b]) \to \mu ((a,b])$.
A: We will only prove here that definition (1) implies definition (2) (this is the easy part).
Suppose that the condition in definition (1) holds. That is, $\mu_n ((a,b]) \to \mu ((a,b])$ for any $a,b \in D$, $a < b$, where $D$ is a dense subset of $R$. Define $\tilde D = \lbrace x \in R:\mu (\lbrace x \rbrace ) = 0 \rbrace$. Then $\tilde D$ is dense in $R$, since its complement is at most countable. For any $a,b \in \tilde D$, $a < b$, and $a',a'',b',b'' \in D$, $a' < b'$, $a'' < b''$, such that $a'' < a < a'$ and $b' < b < b''$, we have
$$
\mu _n ((a',b']) \le \mu _n ((a,b)) \le \mu _n ((a'',b'']).
$$
By assumption, $\mu _n ((a',b']) \to \mu ((a',b'])$ and $\mu _n ((a'',b''] \to \mu ((a'',b''])$ as $n \to \infty$. Then, since $a',a''$ and $b',b''$ can be chosen arbitrarily close to $a$ and $b$, respectively, we can conclude that 
$$
\mu ((a,b)) \le \mathop {\lim \inf }\limits_{n \to \infty } \mu _n ((a,b)) \le \mathop {\lim \sup }\limits_{n \to \infty } \mu _n ((a,b)) \le \mu ([a,b]).
$$
However, by assumption $\mu (\lbrace a \rbrace) = \mu (\lbrace b \rbrace) = 0$, and hence $\lim _{n \to \infty } \mu _n ((a,b)) = \mu ((a,b))$. Thus, the condition in definition (2) is satisfied (with $\tilde D$ playing the role of $D$ in that definition).
It is instructive to consider here the following simple example. As is customary, denote by $\delta_x$ the probability measure concentrated at $x$. Define $\mu_n = \delta_{-1/n}$, $n \geq 1$, and $\mu = \delta_0$. Then, for any $a,b \in R$, $a < b$, we have $\mu _n ((a,b]) \to \mu ((a,b])$, and hence, by definition (1), $\mu_n$ converges vaguely to $\mu$. On the other hand, $\mu _n ((-2,0)) = 1$ for any $n \geq 1$, whereas $\mu ((-2,0)) = 0$. However, if we define $\tilde D$ as above, then $\tilde D = R - \lbrace 0 \rbrace$, and for any $a,b \in \tilde D$, we do have $\mu _n ((a,b)) \to \mu ((a,b))$; hence, $\mu_n$ converges vaguely to $\mu$ also according to definition (2).
Finally, as for the OP's remark (concerning definition (2)) that "the original text just says converges, not mention that converges to $ \mu((a,b))$ which I guess can be added?", it is clear that we are deliberately only given that $\mu_n ((a,b))$ converges -- this is what makes the converse implication (from definition (2) to (1)) a substantially more difficult problem.
A: Try $(a,b] = \bigcap_{c \in D, c > b} (a,c)$ and $(a,b) = \bigcup_{c \in D, c < b} (a,c]$.
